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          <p>近期研究docker，同时也有同事在问，如何使用docker搭建redis集群，我自己也实践下，本机环境是：</p><ul>
<li>操作系统:macOS 10.15</li>
<li>docker:18.09.2</li>
<li>redis:5.0.5</li>
</ul><p>为了能够顺利地使用docker来部署redis集群，我首先本地做redis集群，确保集群的方式是对的，再来到docker上面做redis的集群。</p><h1 id="本地做Redis集群"><a href="#本地做Redis集群" class="headerlink" title="本地做Redis集群"></a>本地做Redis集群</h1><p>本地做redis集群，也是分几个小的步骤，首先配置6个不同redis文件，用于做3个master 3个slaver的集群，然后启动redis server分别启动这个6个redis，最后使用创建集群。</p><h2 id="配置redis文件"><a href="#配置redis文件" class="headerlink" title="配置redis文件"></a>配置redis文件</h2><ul>
<li>首先在任意目录下，创建一个redis配置文件的模板 redis.conf，用于配置redis开启集群模式，代码如下：</li>
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          <h1 id="1-安装nodejs、npm、git"><a href="#1-安装nodejs、npm、git" class="headerlink" title="1. 安装nodejs、npm、git"></a>1. 安装nodejs、npm、git</h1><p>建议安装最新稳定版本的node 和npm，如果版本太久，在搭建过程中，可能会出现一些安装失败的情况。我的版本情况如下：  </p><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br></pre></td><td class="code"><pre><span class="line">localhost:~ hzy$ node -v</span><br><span class="line">v10.16.3</span><br><span class="line">localhost:~ hzy$</span><br><span class="line">localhost:blog-test hzy$ npm -v</span><br><span class="line">6.9.0</span><br><span class="line">localhost:~ hzy$</span><br><span class="line">localhost:~ hzy$</span><br><span class="line">localhost:~ hzy$ git --version</span><br><span class="line">git version 2.20.1</span><br></pre></td></tr></table></figure><h1 id="2-安装hexo"><a href="#2-安装hexo" class="headerlink" title="2. 安装hexo"></a>2. 安装hexo</h1><p>使用npm安装hexo的最新版本，如下:  </p><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br></pre></td><td class="code"><pre><span class="line">localhost:~ hzy$ npm install -g hexo-cli</span><br><span class="line">/Users/hzy/work/repos/npm/bin/hexo -&gt; /Users/hzy/work/repos/npm/lib/node_modules/hexo-cli/bin/hexo</span><br><span class="line">+ hexo-cli@3.0.0</span><br><span class="line">updated 1 package in 5.555s</span><br></pre></td></tr></table></figure>
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          <h1 id="需求分析"><a href="#需求分析" class="headerlink" title="需求分析"></a>需求分析</h1><p>在我们以往做过的各个项目中，用户为了给单据做归档，尤其是根据打印后的纸质单据做归档，往往要求单据号要特定的规则，有的要有流水号，有的要随机号，有的单据需要带前缀，有的要带后缀。举个例子，我们要做一个送货单的单据号生成服务，希望送货单是由前缀D + 日期 + 流水号的方式组成，例如D20190926000001、D20190926000002、D20190926000003。    </p><p>对于各种单据号的生成，其实很多是很类似的，我们需要写一个通用发号机功能，并作为一个公共服务开放出来，可以支持各种发号的规则，同时支持高并发的情况。</p><p>该文章的所有源码请参考如下：<br><a href="https://github.com/hzy2009/id-generator" target="_blank" rel="noopener">https://github.com/hzy2009/id-generator</a></p>
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          <h2 id="事件驱动模型"><a href="#事件驱动模型" class="headerlink" title="事件驱动模型"></a>事件驱动模型</h2><p>事件驱动模型也就是我们常说的观察者模式，或者发布-订阅模型，理解它的几个关键点：</p><ul>
<li>首先是一种对象间的一对多关系，比如微信和微信公众号的关系，微信是观察者，微信公众号是被观擦者</li>
<li>当目标(微信公众号)发生改变，观察者(微信用户)可以接收到改变</li>
<li>观察者如何处理，目标无需干涉，所以就降低耦合度</li>
</ul><h2 id="用户注册的例子"><a href="#用户注册的例子" class="headerlink" title="用户注册的例子"></a>用户注册的例子</h2><p>游戏用户注册成功后，可能会发生这么多事</p><ul>
<li>加积分</li>
<li>发确认邮件</li>
<li>获得游戏大礼包</li>
</ul><p><strong><em>问题</em></strong><br>UserServiceImple类里需要引用创建积分的service，确认邮件的service等，耦合严重，增删功能比较麻烦，也有可能需要调用第三方接口，速度慢，代码维护性差，同时访问数据库多次，用户体验差。    </p>
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          <h2 id="前言"><a href="#前言" class="headerlink" title="前言"></a>前言</h2><p>之前在网上看到过很多关于mysql联合索引最左前缀匹配的文章，自以为就了解了其原理，最近面试时和面试官交流，发现遗漏了些东西，这里自己整理一下这方面的内容。</p><h2 id="最左前缀匹配原则"><a href="#最左前缀匹配原则" class="headerlink" title="最左前缀匹配原则"></a>最左前缀匹配原则</h2><p>在mysql建立联合索引时会遵循最左前缀匹配的原则，即最左优先，在检索数据时从联合索引的最左边开始匹配，示例：<br>对列col1、列col2和列col3建一个联合索引</p><p>KEY test_col1_col2_col3 on test(col1,col2,col3);<br>联合索引 test_col1_col2_col3 实际建立了(col1)、(col1,col2)、(col,col2,col3)三个索引。</p><p>SELECT * FROM test WHERE col1=“1” AND clo2=“2” AND clo4=“4”<br>上面这个查询语句执行时会依照最左前缀匹配原则，检索时会使用索引(col1,col2)进行数据匹配。</p>
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          <p>今天面试的时候，和面试官谈到一个事务级别的问题，后来我自己查了资料，以及自己在Mysql上实践一下，得出一下结论。我操作的Mysql的版本是 MySQL5.6.36+InnoDB</p><h2 id="一、事务的基本要素（ACID）"><a href="#一、事务的基本要素（ACID）" class="headerlink" title="一、事务的基本要素（ACID）"></a>一、事务的基本要素（ACID）</h2><ol>
<li><p>原子性（Atomicity）：事务开始后所有操作，要么全部做完，要么全部不做，不可能停滞在中间环节。事务执行过程中出错，会回滚到事务开始前的状态，所有的操作就像没有发生一样。也就是说事务是一个不可分割的整体，就像化学中学过的原子，是物质构成的基本单位。</p>
</li>
<li><p>一致性（Consistency）：事务开始前和结束后，数据库的完整性约束没有被破坏 。比如A向B转账，不可能A扣了钱，B却没收到。</p>
</li>
<li><p>隔离性（Isolation）：同一时间，只允许一个事务请求同一数据，不同的事务之间彼此没有任何干扰。比如A正在从一张银行卡中取钱，在A取钱的过程结束前，B不能向这张卡转账。</p>
</li>
<li><p>持久性（Durability）：事务完成后，事务对数据库的所有更新将被保存到数据库，不能回滚。</p>
</li>
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          <p>在以软件应用为核心的新一轮企业信息化浪潮中，ERP和协同软件是最主要的两个应用点。协同软件以解决企业资源和系统整合为主要目标，着力于企业管理的规范化以及提高工作效率。毫无疑问，协同软件具有非常广泛的应用前景，但是协同软件市场却并非一马平川，可以说是充满了变数。那么，究竟什么是决定协同软件成败的关键因素呢？这里我总结出4个方面原因。 　　</p><h1 id="1、协同软件向管理咨询靠拢"><a href="#1、协同软件向管理咨询靠拢" class="headerlink" title="1、协同软件向管理咨询靠拢 　　"></a>1、协同软件向管理咨询靠拢 　　</h1><p>协同软件从本质上来说，是解决公司内部管理的问题，是提升公司内部管理水平的一种重要手段。所以协同软件不是单纯的应用软件，而是蕴含着深刻管理思想的应用管理软件。目标决定行为，将协同软件的实施与管理咨询结合起来，这是未来协同软件拥有良好出路的重要前提。 　　</p>
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          <p>给定一个整数类型的数组 nums，请编写一个能够返回数组“中心索引”的方法。  </p><p>我们是这样定义数组中心索引的：数组中心索引的左侧所有元素相加的和等于右侧所有元素相加的和。  </p><p>如果数组不存在中心索引，那么我们应该返回 -1。如果数组有多个中心索引，那么我们应该返回最靠近左边的那一个。  </p><p>示例 1:</p><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br></pre></td><td class="code"><pre><span class="line">输入: </span><br><span class="line">nums = [1, 7, 3, 6, 5, 6]</span><br><span class="line">输出: 3</span><br><span class="line">解释: </span><br><span class="line">索引3 (nums[3] = 6) 的左侧数之和(1 + 7 + 3 = 11)，与右侧数之和(5 + 6 = 11)相等。</span><br><span class="line">同时, 3 也是第一个符合要求的中心索引。</span><br></pre></td></tr></table></figure><p>示例 2:  </p><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br></pre></td><td class="code"><pre><span class="line">输入: </span><br><span class="line">nums = [1, 2, 3]</span><br><span class="line">输出: -1</span><br><span class="line">解释: </span><br><span class="line">数组中不存在满足此条件的中心索引。</span><br></pre></td></tr></table></figure>
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          <p>广度优先搜索（BFS）的一个常见应用是找出从根结点到目标结点的最短路径。  </p>
<p>以下题目：<br>给定一个由 ‘1’（陆地）和 ‘0’（水）组成的的二维网格，计算岛屿的数量。一个岛被水包围，并且它是通过水平方向或垂直方向上相邻的陆地连接而成的。你可以假设网格的四个边均被水包围。  </p>
<p>示例 1:  </p>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br></pre></td><td class="code"><pre><span class="line">输入:</span><br><span class="line">11110</span><br><span class="line">11010</span><br><span class="line">11000</span><br><span class="line">00000</span><br><span class="line"></span><br><span class="line">输出: 1</span><br></pre></td></tr></table></figure>

<p>示例 2:  </p>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br></pre></td><td class="code"><pre><span class="line">输入:</span><br><span class="line">11000</span><br><span class="line">11000</span><br><span class="line">00100</span><br><span class="line">00011</span><br><span class="line"></span><br><span class="line">输出: 3</span><br></pre></td></tr></table></figure>

<p>java代码：</p>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br><span class="line">50</span><br><span class="line">51</span><br><span class="line">52</span><br><span class="line">53</span><br><span class="line">54</span><br><span class="line">55</span><br><span class="line">56</span><br><span class="line">57</span><br><span class="line">58</span><br><span class="line">59</span><br><span class="line">60</span><br><span class="line">61</span><br><span class="line">62</span><br><span class="line">63</span><br><span class="line">64</span><br><span class="line">65</span><br><span class="line">66</span><br><span class="line">67</span><br><span class="line">68</span><br><span class="line">69</span><br><span class="line">70</span><br><span class="line">71</span><br><span class="line">72</span><br><span class="line">73</span><br><span class="line">74</span><br><span class="line">75</span><br><span class="line">76</span><br><span class="line">77</span><br><span class="line">78</span><br><span class="line">79</span><br><span class="line">80</span><br><span class="line">81</span><br><span class="line">82</span><br><span class="line">83</span><br><span class="line">84</span><br><span class="line">85</span><br><span class="line">86</span><br><span class="line">87</span><br><span class="line">88</span><br><span class="line">89</span><br><span class="line">90</span><br><span class="line">91</span><br><span class="line">92</span><br><span class="line">93</span><br><span class="line">94</span><br><span class="line">95</span><br><span class="line">96</span><br><span class="line">97</span><br><span class="line">98</span><br><span class="line">99</span><br><span class="line">100</span><br><span class="line">101</span><br><span class="line">102</span><br><span class="line">103</span><br><span class="line">104</span><br><span class="line">105</span><br><span class="line">106</span><br><span class="line">107</span><br><span class="line">108</span><br><span class="line">109</span><br></pre></td><td class="code"><pre><span class="line">public class NumIslands &#123;</span><br><span class="line">    </span><br><span class="line">    public static int numIslands(char[][] grid) &#123;</span><br><span class="line">        </span><br><span class="line">        int numIsland = 0;</span><br><span class="line">        for(int y=0; y &lt; grid.length; y++)&#123;</span><br><span class="line">            </span><br><span class="line">            for(int x = 0; x&lt;grid[y].length; x++)&#123;</span><br><span class="line">                </span><br><span class="line">                char data = grid[y][x];</span><br><span class="line">                if (data != &apos;1&apos;) &#123;</span><br><span class="line">                    continue;</span><br><span class="line">                &#125;</span><br><span class="line">                </span><br><span class="line">                String rootPostion = y + &quot;-&quot; + x;</span><br><span class="line">                markIsland(rootPostion, grid);</span><br><span class="line">                numIsland++;</span><br><span class="line">            &#125;</span><br><span class="line">            </span><br><span class="line">        &#125;</span><br><span class="line">        return numIsland;</span><br><span class="line">    &#125;</span><br><span class="line">    </span><br><span class="line"></span><br><span class="line">    private static void markIsland(String rootPosition, char[][] grid) &#123;</span><br><span class="line">        Queue&lt;String&gt; root = new LinkedList&lt;&gt;();</span><br><span class="line">        Queue&lt;String&gt; subLand = new LinkedList&lt;&gt;();</span><br><span class="line">        </span><br><span class="line">        Set&lt;String&gt; existPosition = new HashSet&lt;&gt;();</span><br><span class="line">        </span><br><span class="line">        root.offer(rootPosition);</span><br><span class="line">        boolean isEnd = false;</span><br><span class="line">        </span><br><span class="line">        while (!isEnd) &#123;</span><br><span class="line">            if (root.isEmpty() &amp;&amp; !subLand.isEmpty()) &#123;</span><br><span class="line">                root.addAll(subLand);</span><br><span class="line">                subLand.clear();</span><br><span class="line">            &#125;</span><br><span class="line">            </span><br><span class="line">            if (root.isEmpty() &amp;&amp; subLand.isEmpty()) &#123;</span><br><span class="line">                isEnd = true;</span><br><span class="line">                break;</span><br><span class="line">            &#125;</span><br><span class="line">            </span><br><span class="line">            rootPosition = root.poll();</span><br><span class="line">            </span><br><span class="line">            Integer yPosition = Integer.valueOf(rootPosition.split(&quot;-&quot;)[0]);</span><br><span class="line">            Integer xPosition = Integer.valueOf(rootPosition.split(&quot;-&quot;)[1]);</span><br><span class="line">            </span><br><span class="line">            grid[yPosition][xPosition] = &apos;2&apos;;</span><br><span class="line">            </span><br><span class="line">            List&lt;String&gt; subLandArray = getSubPosition(rootPosition, grid);</span><br><span class="line">            if (subLandArray == null || subLandArray.isEmpty()) &#123;</span><br><span class="line">                continue;</span><br><span class="line">            &#125;</span><br><span class="line">            </span><br><span class="line">            for(String tmp : subLandArray)&#123;</span><br><span class="line">                if (existPosition.add(tmp)) &#123;</span><br><span class="line">                    subLand.offer(tmp);</span><br><span class="line">                &#125;</span><br><span class="line">            &#125;</span><br><span class="line">            </span><br><span class="line">        &#125;</span><br><span class="line">        </span><br><span class="line">    &#125;</span><br><span class="line"></span><br><span class="line">    private static List&lt;String&gt; getSubPosition(String rootPosition, char[][] grid) &#123;</span><br><span class="line">        Integer yPosition = Integer.valueOf(rootPosition.split(&quot;-&quot;)[0]);</span><br><span class="line">        Integer xPosition = Integer.valueOf(rootPosition.split(&quot;-&quot;)[1]);</span><br><span class="line">        </span><br><span class="line">        List&lt;String&gt; land = new ArrayList&lt;&gt;();</span><br><span class="line">        </span><br><span class="line">        Integer yNorth = yPosition - 1;</span><br><span class="line">        if (yNorth &gt;= 0 &amp;&amp; grid[yNorth][xPosition] == &apos;1&apos;) &#123;</span><br><span class="line">            land.add(yNorth +&quot;-&quot; + xPosition);</span><br><span class="line">        &#125;</span><br><span class="line">        </span><br><span class="line">        Integer xWest = xPosition - 1;</span><br><span class="line">        if (xWest &gt;=0 &amp;&amp; grid[yPosition][xWest] == &apos;1&apos;) &#123;</span><br><span class="line">            land.add(yPosition + &quot;-&quot; + xWest);</span><br><span class="line">        &#125;</span><br><span class="line">        </span><br><span class="line">        Integer ySouth = yPosition + 1;</span><br><span class="line">        if (ySouth &lt; grid.length &amp;&amp; grid[ySouth][xPosition] == &apos;1&apos;) &#123;</span><br><span class="line">            land.add(ySouth + &quot;-&quot; + xPosition);</span><br><span class="line">        &#125;</span><br><span class="line">        </span><br><span class="line">        Integer xEast = xPosition + 1;</span><br><span class="line">        if (xEast &lt; grid[yPosition].length &amp;&amp; grid[yPosition][xEast] == &apos;1&apos;) &#123;</span><br><span class="line">            land.add(yPosition + &quot;-&quot; + xEast);</span><br><span class="line">        &#125;</span><br><span class="line">        </span><br><span class="line">        return land;</span><br><span class="line">    &#125;</span><br><span class="line"></span><br><span class="line"></span><br><span class="line">    public static void main(String[] args) &#123;</span><br><span class="line">//        char[][] grid = new char[4][5];</span><br><span class="line">//        grid[0] = new char[]&#123;&apos;1&apos;,&apos;1&apos;,&apos;0&apos;,&apos;1&apos;,&apos;0&apos;&#125;;</span><br><span class="line">//        grid[1] = new char[]&#123;&apos;1&apos;,&apos;1&apos;,&apos;0&apos;,&apos;0&apos;,&apos;0&apos;&#125;;</span><br><span class="line">//        grid[2] = new char[]&#123;&apos;0&apos;,&apos;0&apos;,&apos;1&apos;,&apos;0&apos;,&apos;1&apos;&#125;;</span><br><span class="line">//        grid[3] = new char[]&#123;&apos;1&apos;,&apos;0&apos;,&apos;0&apos;,&apos;1&apos;,&apos;0&apos;&#125;;</span><br><span class="line">        char[] grid1 = new char[]&#123;&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;0&apos;,&apos;0&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;0&apos;,&apos;0&apos;,&apos;0&apos;,&apos;0&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;,&apos;1&apos;&#125;;</span><br><span class="line">        char[][] grid = new char[][]&#123;grid1&#125;;</span><br><span class="line">        System.out.println(numIslands(grid));</span><br><span class="line">        </span><br><span class="line">    &#125;</span><br><span class="line"></span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>
        
      
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            <a href="/2019/10/02/开发管理-git管理规范/" class="post-title-link" itemprop="url">开发管理-GIT分支管理</a>
          
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          <p>&emsp;&emsp;分支策略实际上是软件开发中很重要的决策。决定了分支策略，其实就决定了我们开发团队的合作模式，代码的结构，后续的衔接CI/CD的流水线。  </p><p>&emsp;&emsp;考虑选择什么样的分支策略的时候，要把分支策略放到整个持续交付的框架中考虑。配合我们团队模型，采用不同的分支结构，可以达到不同的分支效率。  </p><p>&emsp;&emsp;按照职能划分的分支结构，发布速度是比较慢，可能很多人不太理解为什么这个分支的结构和职能的相关。我们看分支的名称就知道，我们的分支有developer、test、release。其实这些分支分别对应了我们的开发人员、测试人员、运维人员。也就是一般来说，是按照不同的阶段，不同的职能来划分，那么当我们采用这样划分策略的时候，意味着任何一段要交付给客户的代码，到了要发布上线的时候。他必须要经过开发测试到发布，甚至在里面循环好几次.这种情况下，就会出现，不仅人员之间的交接，还有代码下的交接，甚至回滚。如果要我们开发速度提高上去，我们就要选择这种比较优化的模型。</p>
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